Question 1 :
Solve the following equations
(i) sin2 x − 5sin x + 4 = 0
Solution :
Let y = sin x
y2 − 5 y + 4 = 0
(y - 1) (y - 4) = 0
y = 1 and y = 4
sin x = 1 General solution = nπ ± (-1)n y y = π/2 Here n = 2π = 2nπ + (π/2) |
sin x = 4 No solution. |
(ii) 12x3 + 8x = 29x2 − 4
Solution :
12x3 + 8x = 29x2 − 4
12x3 - 29x2 + 8x + 4 = 0
(x - 2) is a factor.The other factors are 12x2 - 5x - 2
12x2 - 5x - 2 = 0
12x2 - 8x + 3x - 2 = 0
4x(3x - 2) + 1(3x - 2) = 0
(4x + 1) (3x - 2) = 0
So, the factors are (x - 2) (4x + 1) (3x - 2)
x - 2 = 0, 4x + 1 = 0, 3x - 2 = 0
x = 2, x = -1/4, x = 2/3
Hence 2, -1/4 and 2/3 are roots of the given polynomial.
Let an xn + ........ + a1x + a0 with an ≠ 0 and, be a polynomial with integer coefficients. If p/q, with ( p, q) =1, is a root of the polynomial, then p is a factor of a0 and q is a factor of an .
Question 2 :
Examine for the rational roots of
(i) 2x3 − x2 − 1 = 0
Solution :
Factor of 1 are ± 1 = p
Factor of 2 are ±1 and ±2 = q
p/q = 1/1, 1/2, -1/1, -1/2
If the given equation will have rational solution, then it must be one of the above given solutions.
1 is the rational number and it is the factor of the given polynomial. 1/2 is not the factor of the given polynomial.
Hence the given polynomial has rational roots.
(ii) x8 − 3x + 1 = 0
Solution :
Factor of 1 are ± 1 (the constant) = p
Factor of 1 are ±1 (coefficient of x8) = q
p/q = 1, -1
If the given equation will have rational solution, then it must be one of the above given solutions.
Hence the given polynomial will not have rational roots.
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